June 2022 Paper 1 Q7
7 In this question the \(x\)- and \(y\)-directions are horizontal and vertically upwards respectively and the origin is on horizontal ground.
A ball is thrown from a point 5 m above the origin with an initial velocity \(\begin{pmatrix}14\\7\end{pmatrix}\mathrm{m\,s^{-1}}\).
| Scheme | Marks | AO |
|---|---|---|
| \(x = 14t\) | B1 | 1.1b |
| \(y = 7t - \frac{1}{2}gt^2 + 5\) | M1 | 1.1b |
| So the position vector is \(\begin{pmatrix}14t\\7t - 4.9t^2 + 5\end{pmatrix}\) | A1 | 2.5 |
| [3] |
Notes
B1: must be \(x = \ldots\) or seen as the first component of the vector.
Do not award for an expression that adds a vector to a scalar
M1: allow without \(+5\), or if \(-5\) seen
Do not award for an expression that adds a vector to a scalar
A1: Must be a single vector.
Accept \(\frac{1}{2}g\) in final answer
Accept \(14t\,\mathbf{i} + \left(7t - 4.9t^2 + 5\right)\mathbf{j}\)
SC1: for \(\begin{pmatrix}7t - 4.9t^2 + 5\\14t\end{pmatrix}\) or \(\begin{pmatrix}14t - 4.9t^2 + 5\\7t\end{pmatrix}\)
| Scheme | Marks | AO |
|---|---|---|
| Lands when \(y = 0\) \(7t - 4.9t^2 + 5 = 0\) | M1 | 3.1b |
| \(t = 1.95\) | A1 | 1.1b |
| gives \(x = 14t = 27.3\) m | B1 | 1.1b |
| [3] |
Notes
M1: Award for correct quadratic or an attempt to find value of \(t\) when their quadratic \(y = 0\)
A1: cao
B1: FT their \(t\) and their linear expression for \(x\)
ISW where candidates find the distance from the point of projection